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CIE 9231 2022 November Paper 13 Q2

A Level / CIE / FP1

CIE 9231 2022 Nov Paper 13 Paper · Question 2

题目

Problem

Prove by mathematical induction that, for all positive integers nn, 72n+97n−507^{2n}+97^n-50 is divisible by 4848.

(6)
题目中文翻译

用数学归纳法证明:对于所有正整数 nn,72n+97n−507^{2n}+97^n-50 都能被 4848 整除。

(6)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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相邻指数项分别带出49与97,与原表达式相减后得到48与96两个倍数。

答题过程

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Let Fn=72n+97n−50F_n=7^{2n}+97^n-50. For n=1n=1,

F1=49+97−50=96=48×2,F_1=49+97-50=96=48\times2,

so the statement holds.

Assume FkF_k is divisible by 4848 for some positive integer kk. Then

Fk+1−Fk= (49−1)72k+(97−1)97k= 48⋅72k+96⋅97k= 48(72k+2⋅97k),\begin{align*} F_{k+1}-F_k =&\,(49-1)7^{2k}+(97-1)97^k\\[4mm] =&\,48\cdot7^{2k}+96\cdot97^k\\[4mm] =&\,48\big(7^{2k}+2\cdot97^k\big), \end{align*}

which is divisible by 4848. The hypothesis therefore implies that Fk+1F_{k+1} is divisible by 4848.

Since the statement holds for n=1n=1 and truth for n=kn=k implies truth for n=k+1n=k+1, the result follows by induction for every positive integer nn.